Forum:Why does Madore's Psi function get stuck at zeta 0 ?
I don't fully understand the reason why \psi(\alpha)=\zeta_0 when \zeta_0 \leq\alpha\leq\Omega , and though I asked, it still seems arbitrary to me. Consider that if \psi(0)=\epsilon_0 , \psi(1)=\epsilon_1 , \psi(2)=\epsilon_2 , and so on, then in general, \psi(\alpha)=\epsilon_{\alpha} . So why should \psi(\zeta_0 +1) be equal to \zeta_0 instead of \epsilon_{\zeta_0 +1} ? Also, if \psi(\zeta_0)=\zeta_0 , then logically \psi(\zeta_0 +1) could be seen to be equal to {\zeta_0}^ }}} , according to the definition of the psi function. Therefore, would \epsilon_{\zeta_0 +1}={\zeta_0}^ }}} ? Clearly this cannot be so, but I would like to know where I went wrong, and how, in complete and understandable terms, the Psi function gets stuck at \zeta_0 , and why \psi(\zeta_0 +1) should behave any differently than \psi(\Omega +1) . Edwin Shade (talk) 01:24, November 13, 2017 (UTC) : The reason is actually quite simple. ψ(α) is defined as the smallest ordinal which can't constructed by using: : (1) The ordinals 0, 1, ω and Ω : (2) Ordinal arithmetics (addition, multiplication and exponentiation) : (3) The function ψ(β), which can only be used for previously constructed ordinals β<α : The bolded section is the important part here. We are not allowed to use the function ψ(β) on every ordinal less than α. We are required to build the ordinal β from the above building blocks as well, in order use it. : For example, when analyzing ψ(ε₀+1), we find that it is equal to: : ' ψ(ε₀+1) = ψ(ε₀)^ψ(ε₀)^ψ(ε₀)^... = εε₀+1' : We are only allowed to do this because ε₀ itself can be written with the basic building blocks (0,1,ω,Ω,ψ): : ' ψ(ε₀+1) = ψ(ψ(0))^ψ(ψ(0))^ψ(ψ(0))^... = εε₀+1' : It is quite easy to show that this can be done with any ordinal below ζ₀, which is why ψ(a)=εa for such ordinals. : But at ζ₀ itself things get interesting: Since ζ₀ is a fixed point of the ε numbers, the smallest ordinal for which ψ(x)=ζ₀ is ζ₀ itself. So we have a catch 22: We can't build ζ₀ with the ψ function unless we already have ζ₀ itself at our disposal. So if you try to write: : ψ(ζ₀+1) = ψ(ζ₀)^ψ(ζ₀)^ψ(ζ₀)^... (this is wrong) : You can't justify this equation, because you aren't allowed to use "ζ₀" in your construction. : Indeed, since ζ₀ cannot be constructed from 0's and 1's and ω's and ψ's, it will be forever unreachable. This is why we have: : ' ψ(x)=ζ₀ for ζ₀ ≤ x ≤ Ω'. : So, how does the function gets "unstuck" at Ω? : Well, remember that we are allowed to use Ω itself in our constructions. Therefore, we can actually construct the ordinal ζ₀ by using Ω: : ' ζ₀ = ψ(Ω)' : This is a perfectly valid construction. The reason it didn't help us before, is that we aren't allowed to use ψ(a) when calculating ψ(b) for bΩ. : And so we have: : ' ψ(Ω+1) = ψ(Ω)^ψ(Ω)^ψ(Ω)^... = ζ₀^ζ₀^ζ₀^... = εζ₀+1' : And from here we continue normally. : Note that from now on we can use the ordinal ζ₀=ψ(Ω) freely. So when we get to: : ' ψ(Ω+ζ₀) = εζ₀×2' : We don't have a similar problem. Indeed, it can be shown that: : ' ψ(Ω+a) = εζ₀+a for any a ≤ ζ₁' : At ζ₁ we get stuck again (for very similar reasons): : ' ψ(Ω+ζ₁) = ζ₁' : Until Ω comes to the rescue once again, with the representation: : ' ζ₁ = ψ(Ω+Ω) = ψ(Ω×2) ' : ' PsiCubed2 (talk) 03:40, November 13, 2017 (UTC)' \(C(\zeta_0+1)\) doesn't contain \(\zeta_0\). From \(0,\ 1,\ \omega\), we can apply addition, multiplication, exponentiation and \(\psi()\), then we can have \(\psi(0)=\varepsilon_0\), \(\psi(\psi(0))=\varepsilon_{\varepsilon_0}\), and so on. But we can't get \(\zeta_0\) from those things. {hyp/^,cos} (talk) 02:19, November 13, 2017 (UTC) So if I understand correctly then, \Omega is really a symbol telling us to find the first ordinal not accessible through the system of functions we've developed so far, rather than being equal to a real ordinal ? Edwin Shade (talk) 03:36, November 20, 2017 (UTC) : It's both. : The purpose of Ω is exactly what you said. : The mechanism by which this works is that Ω is a very large ordinal (usually we set Ω=ω₁) that we are free to "fetch" without constructing it first. : And yes, it is an ordinal just like any other. That's why we can have things like Ω+1 or Ω2 or even ΩΩ. The fact that we can do ordinal arithmetic with Ω's is a crucial part of the system. PsiCubed2 (talk) 05:06, November 20, 2017 (UTC)